A thermally insulated pot has 150 g ice at temperature ` 0^(@)C`. How much steam of ` 100^(@) C ` has to be mixed to it , so that water of temperature ` 50^(@)C` will be obtained ? ( Given : Latent heat of melting of ice = 80 cal/g, latent of vaporization of water = 540 cal/g, spcific heat of water = ` 1 cal// g.^(@)C)`

Explanation

To solve the problem step by step, we will calculate the amount of heat required to convert the ice into water at 50°C and then determine how much steam at 100°C is needed to provide that heat.

### Step 1: Calculate the heat required to convert ice at 0°C to water at 0°C
The formula to calculate the heat required to melt ice is:
\[ Q_1 = m \times L_f \]
Where:
- \( m = 150 \, \text{g} \) (mass of ice)
- \( L_f = 80 \, \text{cal/g} \) (latent heat of fusion)
...